Symfony 3.2 绑定 post 值以在 REST api 中形成

Question

我使用以下 libraries/technologies:

JMSSerializer、FOSRestBundle + Symfony 3.2 + PHP 7.1

当我尝试向我的 POST 端点发出 POST 请求时，我无法使用表单。

文件：

Country.php -> POPO 实体

CountryType.php

<?php

declare(strict_types = 1);
namespace AppBundle\Form;

use AppBundle\Model\Entity\Country;
use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\Extension\Core\Type\IntegerType;
use Symfony\Component\Form\Extension\Core\Type\TextType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolver;

class CountryType extends AbstractType
{
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder->add('name' , TextType::class)
            ->add('iso_alpha_2_code', TextType::class)
            ->add('iso_alpha_3_code', TextType::class)
            ->add('is_numeric_code', IntegerType::class);
    }

    public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults([
            'data_class' => Country::class,
            'csrf_protection' => false,
        ]);
    }

    /**
     * @inheritdoc
     */
    public function getName() : string
    {
        return '';
    }
}

CountryController.php postAction

public function postAction(Request $request)
    {
        $country = new Country();
        $form = $this->createForm(CountryType::class,  $country);
        $form->handleRequest($request);

        if ($form->isSubmitted() && $form->isValid()) {
            die('ok');
            // TODO INSERT DATA then redirect
            return $this->routeRedirectView('get_country', ['id' => $country->getId()]);
        }

        return $this->get('fos_rest.view_handler')->handle(View::create($form));
    }

问题是它不会进入 if 块，因为 isSubmited() 和 isValid() 方法 return 都是假的。当我调用 $form->getData() 时，它 returns

CountryController.php on line 40: Country {#306 -id: null -name: null -isoAlpha2Code: null -isoAlpha3Code: null -isNumericCode: null }

我提出的要求：

你能告诉我我做错了什么吗？

Answer 1

这就是我在使用 FOSRest 包时提交表单的方式，我希望这能让您朝着正确的方向前进。这是一个示例注册操作。您会注意到您必须手动提交表单

$form->submit($request->request->all()); 然后检查是否有效

public function postRegistrationAction(Request $request){
    $form = $this->createForm(UserType::class, null, [
        'csrf_protection' => false,
    ]);
    $form->submit($request->request->all());
    if (!$form->isValid()) {
        return $form;
    }
    /**
     * @var $user User
     */
    $user = $form->getData();
    $em = $this->getDoctrine()->getManager();

    //Set the user role
    $user->setRoles(array('ROLE_USER'));

    //Encode the password
    $password = $request->request->get('password');
    $encodedPassword = $this->get('security.password_encoder')->encodePassword($user, $password['first']);
    $user->setPassword($encodedPassword);

    $token = $this->get('lexik_jwt_authentication.encoder')->encode(['username' => $user->getUsername(), 'role' => $user->getRoles(), 'name' => $user->getName()]);

    $em->persist($user);
    $em->flush();
    $view = FOSView::create();
    $view
        ->setData(['token' => $token])
        ->setStatusCode(200);
    return $view;
}

Answer 2

我能够使用以下行获取数据

    $user = new User;
    $form = $this->createForm(UserType::class, $user);
    $form->submit($request->request->all());
    if($form->isValid()){
       //persist or do anything
       return $user;
    }
    return 'There were an error in your form';

Symfony 3.2 绑定 post 值以在 REST api 中形成

Symfony 3.2 binding post values to form in REST api

php

symfony

fosrestbundle

jmsserializerbundle

symfony-3.2